Problem: Simplify; express your answer in exponential form. Assume $x\neq 0, r\neq 0$. $\dfrac{{(x^{-3})^{-2}}}{{(x^{2}r^{-4})^{5}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${x^{-3}}$ to the exponent ${-2}$ . Now ${-3 \times -2 = 6}$ , so ${(x^{-3})^{-2} = x^{6}}$ In the denominator, we can use the distributive property of exponents. ${(x^{2}r^{-4})^{5} = (x^{2})^{5}(r^{-4})^{5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(x^{-3})^{-2}}}{{(x^{2}r^{-4})^{5}}} = \dfrac{{x^{6}}}{{x^{10}r^{-20}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{6}}}{{x^{10}r^{-20}}} = \dfrac{{x^{6}}}{{x^{10}}} \cdot \dfrac{{1}}{{r^{-20}}} = x^{{6} - {10}} \cdot r^{- {(-20)}} = x^{-4}r^{20}$.